Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-10, sketch the region \(R\) bounded by the graphs of the given equations, and show a typical vertical slice. Then find the volume of the solid generated by revolving \(R\) about the \(x\)-axis. $$ y=x^{3}, x=3, y=0 $$

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{2187 \pi}{7} \).

Step 1: Identify the Bounded Region

First, let's describe the region bounded by the equations. The graph of the function is given by \( y = x^3 \), and the bounds are defined by \( x = 3 \) and \( y = 0 \). The region \( R \) is to the right of the curve \( y = x^3 \), below the line \( x = 3 \), and above the line \( y = 0 \).

Step 2: Sketch the Region

Now, we sketch the region based on the described bounds. The region of interest is the area under the curve \( y = x^3 \) from \( x = 0 \) to \( x = 3 \). The region is a 'wedge' between the \( y \)-axis and the line \( x = 3 \).

Step 3: Consider a Typical Vertical Slice

Consider a slice of the region \( R \) at a point \( x \). The height of this slice is \( y = x^3 \), and the slice has a small width of \( dx \).

Step 4: Set Up the Expression for Volume

The volume of the solid generated by revolving slice about the \( x \)-axis is approximated by the volume of a thin disk. The volume of one such disk is given by:\[ dV = \pi (x^3)^2 \cdot dx = \pi x^6 \cdot dx \]To find the total volume, integrate this expression from \( x = 0 \) to \( x = 3 \).

Step 5: Calculate the Integral

Perform the integration:\[ V = \int_{0}^{3} \pi x^6 \, dx = \pi \int_{0}^{3} x^6 \, dx \]Find the antiderivative of \( x^6 \), which is \( \frac{x^7}{7} \).

Step 6: Evaluate the Integral at the Bounds

Substitute the bounds into the antiderivative:\[ V = \pi \left[ \frac{x^7}{7} \right]_{0}^{3} = \pi \left( \frac{3^7}{7} - \frac{0^7}{7} \right) \]Simplify the expression:\[ V = \pi \left( \frac{2187}{7} \right) \]

Step 7: Finalize the Volume

Calculate the volume:\[ V = \frac{2187 \pi}{7} \] This is the volume of the solid generated by revolving the region about the \( x \)-axis.

Key Concepts

Volume of Solids of Revolution

The concept of finding the volume of solids of revolution involves visualizing a plane region rotating around a given axis. By doing so, the region forms a three-dimensional object, which we can understand as a solid of revolution. In our exercise, we're rotating the bounded region by the successful equations around the x-axis.
The basic strategy is to think of the volume as a collection of infinitely thin disks stacked along the axis about which rotation occurs. Each disk contributes a small volume, collectively forming the whole structure. The volume of each disk is calculated using the radius of the disk, which, in this case, is determined by the distance from the x-axis to the curve.
This approach is particularly useful in calculus as it translates the visual, geometric task of computing volume into a flexible form suitable for integration, allowing us to apply algebraic rules to tractable mathematical expressions.

Integration

Integration is a fundamental tool in calculus used to compute areas under curves, among other applications. In the context of finding volumes of solids of revolution, integration allows us to sum up the volumes of many thin slices.
In our example problem, once we have verbalized a single slice's volume formula as a tiny disk, integration becomes the calculated method for adding up the infinite number of these disks between the limits of the bounded region (from the start to the end of the curve).
More specifically, the definite integral is used because it efficiently accounts for all these small volumes between two bounds. Integration becomes indispensable to moving from a single slice to a full region's volume, providing a complete solution of a complex geometric problem with algebraic elegance.

Definite Integrals

Definite integrals are integrals with clearly specified limits, which in effect restrict the region of integration to between two points. In the exercise, the definite integral helps effectively capture the entire volume of the rotated solid between the x-values of 0 and 3.
The process involves evaluating the integral of the volume formula from the initial bound to the terminal bound. Here, the region under consideration is precise, ensuring the calculated result is accurate for just the part of the space revolved around the axis.
Moreover, definite integrals simplify the calculation as their evaluation encompasses the bounds substitution into the antiderivative, ultimately lending exactness to our volume calculation. This method showcases how calculus can empower us to capture complicated physical phenomena in simple numerical terms.